Tensor decomposition I came across what a Physicist called "decomposing a tensor with respect to a congruence", something I simply cannot grasp. I searched a lot and I couldn't find any reference on that. I know that "congruence" is the name Physicists use to refer to the integral lines of a vector field on a manifold, but I have no idea what it means to decompose something with respect to them. 
Just as a curiosity, the "decomposition" of the electromagnetic tensor "with respect to the congruence of observers $V^\mu$", whatever that means, is
$$F^{\mu \nu} = \frac{1}{c} \left( E^{[\mu}V^{\nu]} - \eta^{\mu \nu}_{\;\;\alpha \beta} B^\alpha V^\beta \right) ,$$
where $E^\mu V_\mu = B^\mu V_\mu = 0$, and nothing was said about $\eta$, so I actually don't know what it is. 
The answer doesn't need to explain why $F$ is actually decomposed this way. Just some reference on the subject and an explanation of what it is are enough.
P.S.: This was said in a General Relativity lecture, that's why I'm tagging it like GR.
 A: In the following, $(\mathcal{M},g,\epsilon)$ is an oriented Lorentz manifold $\mathcal{M}$ with metric tensor $g_{ab}$ and volume form $\epsilon=\sqrt{-\det g}\,\mathrm{d}x^0\wedge\dots\wedge\mathrm{d}x^3$.
Theorem. Given an antisymmetric tensor $A_{ab}$ and a unit timelike vector $u^a$ there exist unique vectors $q^a,b^a$ such that $$A_{ab}=u_aq_b-q_au_b+\epsilon_{abcd}u^cb^d$$
where $\epsilon$ is the canonical volume form on the manifold with the conditions that $q_au^a=b_au^a=0$.
Proof. We will prove this at a point $p\in\mathcal{M}$ using Riemann normal coordinates. The result can be extended pointwise to all of spacetime. 
We set $q_a:=A_{ab}u^b$ and by antisymmetry of $A_{ab}$ it is clear that $q_au^a=0$. Next, define $$B_{ab}:=A_{ab}-u_aq_b+q_au_b.$$ Note that $B_{ab}u^b=0$. Let $u^a,X^a,Y^a,Z^a$ be an orthonormal basis on $T_p\mathcal{M}$. Let us define at $p$ the three numbers 
$$b^1:=B_{ab}Y^aZ^b,b^2:=B_{ab}Z^aX^b,b^3:=B_{ab}X^aY^b$$
and introduce the vector $$b^a:=b^1X^a+b^2Y^a+b^3Z^a$$
in the subspace of $T_p\mathcal{M}$ orthogonal to $u^a$ (henceforth known as $T'_p\mathcal{M}$). It is clear that $u_ab^a=0$. For two vectors $v^a,w^a\in T'_p\mathcal{M}$ one may verify that $B_{ab}v^aw^b$ is the triple scalar product of the vectors $v^a,w^a,b^a$ in the three-dimensional $T'_p\mathcal{M}$ if one assumes a Euclidean metric ($\vec v\cdot(\vec w\times \vec b)$ with the standard dot and cross products on $\mathbb{R}^3$). This is possible if one works in Riemann normal coordinates. Thus we have an antisymmetric tensor $e_{abc}$ for which $B_{ab}v^aw^b=e_{abc}b^av^bw^c$. Using our basis and Riemann normal coordinates, it is easy to verify that $e_{abc}=u^d\epsilon_{dabc}$ (because $u^\mu=(-1,0,0,0)$ and $g_{\mu\nu}\lvert_p=\eta_{\mu\nu}$). We then have $$B_{ab}v^aw^b=\epsilon_{abcd}u^ab^bv^cw^d=(A_{ab}-u_aq_b+q_au_b)v^aw^b.$$ Uniqueness of $q^a$ follows from calculating $A_{ab}v^au^b=q_av^a$ for some $v^a\in T_p\mathcal{M}$ and nothing that the inner product is nondegenerate. Uniqueness of $b^a$ follows from nondegenerateness of the volume form on $T'_p\mathcal{M}$. $\quad\Box$
The claim in the OP follows by identifying $E^a=-\tfrac{1}{2}q^a$ and $B^a=b^a$.
